In aqueous alkaline solution two electron reduction of HO_(2)^(-) give

1.	In aqueous alkaline solution two electron reduction of HO_(2)^(-) gives
Answer» `HO^(-)` `H_(2)O` `O_(2)` `O_(2)^(-)` Solution :`HO_(2)^(-)rarrOH^(-)` O.N of each O atom in `HO_(2)^(-)` is -1 while that in `OH^(-)` is -2 in other WORDS total O.N of two O atoms in`HO_(2)^(-)` is -2 while that of two O atoms in `OH^(-)` -4 THEREFORE to balance O atoms on EITHER side add `2E^(-)` to lL.H.S and one doing so we have `HO_(2)^(-)+2e^(-)rarr 2OH^(-)` Since the reaction is occuring in basic medium therefore to balance H atoms add one `H_(2)O` molecules to L.H.S and one `KO^(-)` ion to R.H.Sof eqn (ii) Thus the balanced EQUATION is `HO_(2)^(-)+H_(2)O+2e^(-)rarr3OH^(-)`

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